# Power Electronics Questions and Answers – Quality of Inverters

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This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “Quality of Inverters”.

1. If Vr is the rms value of the inverter output voltage and V1 is the rms value of the fundamental component, then the total harmonic distortion (THD) is given by
a) Vr/V1
b) (Vr + V1r
c) [ (Vr/V1)2 – 1 ]1/2
d) [ (Vr/V1)1/2 + 1 ]2

Explanation: THD is the ratio of rms value of all the harmonic components to the rms value of the fundamental component.
Rms value of all the harmonic components VH = (Vr2 – V12)1/2
THD = (Vr2 – V12)1/2/ V1.

2. The distortion factor (μ) is the ratio of
a) total rms output voltage to fundamental rms output voltage
b) fundamental rms output voltage to fundamental average output voltage
c) total rms output voltage to rms value of all the harmonic components
d) fundamental rms output voltage to total rms output voltage

Explanation: μ = V1/Vr.

3. A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the fundamental power delivered to the load.
a) 1.07 W
b) 107.2 W
c) 1.07 kW
d) 107.2 kW

Explanation: The fundamental power is the power delivered to the load due to the fundamental components of voltage and current.
V1 (rms) = 2Vs/√2π = 103.552 V
I1 (rms) = V1/R = 10.3552 A
P = (I1)2 x R = 1072.3 W = 1.0723 kW.

4. A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the average current through each SCR inverter switch.
a) 11.5 A
b) 5.75 A
c) 23 A
d) none of the mentioned

Explanation: Peak current through each SCR = Vs/2R = 11.5 A
As each SCR would conduct for 180° of the total 360° cycle, average current = peak current/2 = 11.5/2 = 5.75 A.

5. Find the distortion factor (μ), for a single phase half wave bridge inverter with dc source Vs = 1 kV.
a) 0.87
b) 1
c) 0.9
d) 0.7

Explanation: μ = V1/Vr
V1 = rms value of fundamental component = 2Vs/π√2
Vr = Total rms output voltage = Vs/2
μ = (2Vs/π√2) x (2/Vs) = 2√2/π = 0.9.

6. A single phase inverter gives rms value of output voltage as 115 V and the fundamental output voltage of as 103.5 V. Find the THD (Total Harmonic Distortion).
a) 0.4 %
b) 40.8 %
c) 48.3 %
d) 4.83 %

Explanation:
Vr = 115 V and V1 = 103.5 V
THD =[ (Vr2 – V12)1/2/ V1 ] x 100 %.

7. What would be the harmonic factor of lowest order harmonic in case of a half wave bridge inverter?
a) 1/1
b) 1/3
c) 1/2
d) Insufficient data

Explanation: Let Vs be the input voltage.
The 3rd harmonic is the lowest order harmonic.
Vo (rms) = 2Vs/(n x π x √2)
For fundamental component n = 1 and for the lowest order harmonic n = 3
ρ3 = V3/V1 = 1/3 or 33.33 %.

8. A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the rms output voltage and the peak reverse blocking voltage of each transistor.
a) 30 V, 60 V
b) 30 V, 30 V
c) 60 V, 60 V
d) 60 V, 30 V

Explanation: Rms output voltage = PIV of each transistor = Vs. Vs = 60 V.

9. A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the fundamental frequency output power.
a) 1200 W
b) 856 W
c) 972 W
d) 760 W

Explanation: V1 (rms) = 4Vs/π√2 = 54.02 V
P = (V1)2/R = 972.72 W.

10. Determine the distortion factor (μ) for a full bridge inverter with supply Vs = 60 V.
a) 0.8
b) 0.7
c) 0.9
d) 1

Explanation: V1 (rms) = 4Vs/π√2 = 54.02 V
rms value of output voltage Vr = Vs = 60 V.
μ = V1/Vr = 54.02/60 = 0.9.

Sanfoundry Global Education & Learning Series – Power Electronics.

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